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3 years ago

How can you use Hess’s law to determine a reaction’s enthalpy?

1 answer:

3 0

Hess law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out.

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Plz help guys ASAP! Thanks in advance

MariettaO [177]

Answer:

3.6 moles

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂ to produce 2 moles NH₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂.

Therefore, 3.2 moles of N₂ will react with = 3.2 × 3 = 9.6 moles of H₂.

From the calculation made above, we can see that it will take a higher amount (i.e 9.6 moles) of H₂ than what was given (i.e 5.4 moles) to react completely with 3.2 moles of N₂.

Therefore, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we the greatest quantity of ammonia, NH₃ produced from the reaction.

In this case, the limiting reactant will be use because all of it is consumed in the reaction.

The limiting reactant is H₂ and greatest quantity of ammonia, NH₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, 5.4 of H₂ will react to produce = (5.4 × 2)/3 = 3.6 moles of NH₃

Thus, the greatest quantity of ammonia, NH₃ produced from the reaction is 3.6 moles

4 0

3 years ago

What are the 3 primary consumers and what are the 3 secondary consumers

goldenfox [79]

Primary:
Grasshopper
Mouse
Grass

Secondary:
Hawk
Snake
Coyote

8 0

3 years ago

The boiling point of methanol is 64 7°C. Its melting point is -976°C. At room temperature (-25°C), methanol is in which state?

marissa [1.9K]

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.

The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.

At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.

At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.

You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults

3 0

2 years ago

Read 2 more answers

Identify which of the following two reactions you would expect to occur more rapidly: (1) addition of HBr to 2-methyl-2-pentene

arlik [135]

Answer:

(1) addition of HBr to 2-methyl-2-pentene

Explanation:

In this case, we will have the formation of a carbocation for each molecule. For molecule 1 we will have a tertiary carbocation and for molecule 2 we will have a secondary carbocation.

Therefore the most stable carbocation is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)

3 0

3 years ago

Someone answer the limited regent...

OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

Chemistry - Stoichiometry

Using Dimensional Analysis

Chemistry - Atomic Structure

Reading a Periodic Table

Explanation:

Step 1: Define

[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given] 10 g Cu

Step 2: Identify Conversions

[RxN] 1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

Step 3: Stoichiometry

$10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} )$ = 16.974 g Ag

Step 4: Check

We are given 1 sig fig. Follow sig fig rules and round.

16.974 g Ag ≈ 20 g Ag

6 0

3 years ago